The Bessel identity behind Rudin 8.11-8.12: the exact mean-square error of a linear combination

The Bessel identity (Rudin, Theorems 8.11 and 8.12). Let $\{\varphi_m\}$ be an orthonormal system on $[a,b]$, $a \le b$: $\int_a^b \varphi_n \overline{\varphi_m} = 0$ for $n \ne m$ and $\int_a^b |\varphi_n|^2 = 1$. Let $f$ be square-integrable on $[a,b]$ and let

be its Fourier coefficients relative to the system. Then for every $n$ and every choice of complex numbers $\gamma_0,\dots,\gamma_{n-1}$,

This single identity is the computation behind both of Rudin's theorems. Taking $\gamma_m = c_m$ kills the last sum and shows that the partial sum of the Fourier series is the orthogonal projection of $f$ onto the span of $\varphi_0,\dots,\varphi_{n-1}$: it is the best mean-square approximation, since the last sum is nonnegative for every other choice of $\gamma$ (Theorem 8.11). Since the left-hand side is nonnegative, taking $\gamma = c$ also gives $\sum_{m<n}|c_m|^2 \le \int_a^b \|f\|^2$, which is Bessel's inequality (Theorem 8.12). The identity moreover quantifies the defect: the mean-square error of an arbitrary approximation exceeds the optimal one by exactly $\sum_{m<n} |\gamma_m - c_m|^2$.

Formalization notes. Integrability hypotheses on $f$ are genuinely needed, not decoration: the interval integral of a non-integrable function is $0$ by convention, and without them the statement is false (on $[0,1]$ with $\varphi_m(x) = e^{2\pi i m x}$ and $f(x) = x^{-3/4}$ one has $c_0 = 4$ while $\int_0^1 \|f\|^2$ evaluates to $0$). The hypothesis hf supplies the measurability of $f$ and hf2 its square-integrability. Similarly some measurability of the system is required, supplied by hφint; no separate square-integrability hypothesis on the $\varphi_m$ is needed, because $\int_a^b |\varphi_m|^2 = 1 \ne 0$ already forces $|\varphi_m|^2$ to be integrable.